Figure shows position and velocities of two particles moving under mutual gravitational attraction in space at time t=0. The position of centre of mass after one second is

Options

A. x = 4 m
B. x = 6 m
C. x = 8 m
D. x = 10 m

Answer

D

Explanation

As `vecF_("ext") = 0 , therefore veca_(CM) =0 `
`vecv_(CM) = (2 xx 8 hati - 3 xx 2hati)/(2 + 3) = 2hati , X_(CM) = (2 xx 2 + 3 xx 12)/(2 + 3) = 8`
As `vecv_(CM)` is constant , so centre of mass will move 2m in one second
`therefore` x = 8 + 2 = 10 m